Solvability by Radicals and the Quintic

The quadratic formula, and its harder cousins for the cubic and quartic, express roots using field operations and root extractions. Galois's theorem explains exactly when such a formula exists: the roots of $f$ are expressible in radicals if and only if the Galois group of $f$ is a solvable group. The word solvable for groups is not a coincidence — it was named for this. Because $S_5$ is not solvable, the general quintic has no radical formula, and a specific quintic over $\mathbb{Q}$ with group $S_5$ exhibits a polynomial whose roots provably cannot be written in radicals. We work over a field of characteristic $0$ throughout.

Radical extensions

Solving by radicals means climbing a tower where each step adjoins a root of an element already built.

The chain formalizes a succession of field operations and root extractions. For example, the constructible number for the regular $17$-gon lies in a tower of four square-root extensions, which is why the $17$-gon is constructible by straightedge and compass — only square roots appear.

Cyclic extensions are radical extensions

The link between radicals and groups runs through cyclic extensions, once the base field has enough roots of unity. Roots of unity are themselves radicals (${\zeta }_n=\sqrt[n]{1}$), so they can always be adjoined at the start of any radical tower.

An automorphism $\sigma$ sends $\sqrt[n]{a}$ to ${\zeta }_{\sigma }\sqrt[n]{a}$ for some root of unity ${\zeta }_{\sigma }$, and $\sigma \mapsto {\zeta }_{\sigma }$ is an injective homomorphism into the cyclic group ${\mu }_n$, so the Galois group is cyclic.¹ The converse recovers a radical from a cyclic extension using a Lagrange resolvent.

Applying $\sigma$ multiplies the resolvent by ${\zeta }^{-1}$, so $\sigma ((\beta ,\zeta {)}^n)=(\beta ,\zeta {)}^n$, which places $(\beta ,\zeta {)}^n=a$ in $F$. By independence of characters some $\beta$ gives a nonzero resolvent, and it lies in no proper subfield, so $K=F((\beta ,\zeta ))=F(\sqrt[n]{a})$.

These two propositions are the basis of Kummer theory: with roots of unity present, cyclic and simple-radical extensions coincide.²

Galois's theorem

A solvable group is one with a subnormal series

This is the shape of the Galois-group side of a radical tower.

The two directions each translate one tower into the other.

• Solvable by radicals $\Rightarrow$ solvable group. A radical tower can be enlarged to a Galois radical tower whose successive extensions are cyclic (adjoin roots of unity, take the Galois closure). The corresponding subgroups $G_i=\mathrm{Gal}(L/K_i)$ form a chain with cyclic quotients $G_i/G_{i+1}\cong \mathrm{Gal}(K_{i+1}/K_i)$, so $G$ is solvable. The Galois group of $f$ is a quotient of $G$, hence solvable.
• Solvable group $\Rightarrow$ solvable by radicals. Take a subnormal series with cyclic quotients for $G=\mathrm{Gal}(K/F)$ and its fixed fields $F=K_0\subseteq \cdots \subseteq K_s=K$. After adjoining the needed roots of unity, each $K_{i+1}/K_i$ is cyclic, hence a simple radical extension. So every root of $f$ lies in a radical tower.³

Degrees two through four are solvable

For low degree the symmetric group is solvable, so a radical formula must exist. The composition series exhibit the cyclic quotients that Galois's theorem converts into radicals.

For the cubic $y^3+py+q$, the series $1⊴A_3⊴S_3$ predicts adjoining a cube root of unity and forming Lagrange resolvents. With $\rho$ a primitive cube root of unity and roots $\alpha ,\beta ,\gamma$, the resolvents ${\theta }_1=\alpha +\rho \beta +{\rho }^2\gamma$ and ${\theta }_2=\alpha +{\rho }^2\beta +\rho \gamma$ satisfy ${\theta }_1^3,{\theta }_2^3\in \mathbb{Q}(\sqrt{D},\rho )$, yielding Cardano's formula: with $A=\sqrt[3]{-\frac{27}{2}q+\frac{3}{2}\sqrt{-3D}}$ and $B$ its partner chosen so $AB=-3p$, the roots are $\frac{1}{3}(A+B)$ and its $\rho$-twists. The quartic reduces to its resolvent cubic: the quotient $S_4/V\cong S_3$ is the cubic's group, and the final $V$ step contributes the two square roots that solve for the four roots. This is Ferrari's method.

The quintic is not solvable

At degree $5$ the pattern stops, because the alternating group becomes simple.

Simplicity blocks the descent: the only normal subgroups of $A_5$ are $1$ and $A_5$ itself, and $A_5$ is non-abelian, so any attempt at a subnormal series with cyclic quotients stalls at $A_5$ and never reaches $1$. The derived subgroup $[A_5,A_5]=A_5$, so the derived series is constant. Contrast $S_4$, whose derived series $S_4\supseteq A_4\supseteq V\supseteq 1$ terminates.

Combining with the fact that the generic degree-$n$ polynomial has group $S_n$ gives the classical impossibility result.

An explicit unsolvable quintic

The general result concerns the generic polynomial; a specific rational quintic with a non-solvable Galois group provides an explicit example.

The argument combines irreducibility, a cycle count, and a real-root count.

• A $5$-cycle. $f$ is Eisenstein at $3$, hence irreducible, so the splitting field has degree divisible by $5$ and the Galois group $G\le S_5$ has order divisible by $5$. An element of order $5$ in $S_5$ is a $5$-cycle, so $G$ contains one.
• A transposition. Evaluating gives $f(-2)=-17$, $f(0)=3$, $f(1)=-2$, $f(2)=23$, so $f$ has real roots in $(-2,0)$, $(0,1)$, $(1,2)$. The derivative $f^{\prime }(x)=5x^4-6$ has only two real zeros, so $f$ has exactly three real roots and one conjugate pair of complex roots. Complex conjugation restricts to an automorphism that fixes the three real roots and swaps the two complex ones — a transposition in $G$.

A transitive subgroup of $S_5$ that contains a transposition and a $5$-cycle is all of $S_5$: the $5$-cycle and one transposition generate $S_5$. Since $S_5$ is not solvable, the roots of $x^5-6x+3$ cannot be written in radicals.⁴ The same construction prescribing factorization types modulo small primes builds quintics with group $S_5$ at will, and the reduction-mod-$p$ criterion — cycle types from the factorization of $f\mathrm{mod}p$ — identifies them.

Three classical impossibilities

Galois theory converts three classical impossibilities into statements about groups. The non-constructibility of doubling the cube and trisecting the angle is a degree obstruction; the constructible polygons are governed by cyclotomic Galois groups; and the quintic's lack of a formula is the non-solvability of $S_5$. In each case a long-standing question about numbers is settled by translating it into the finite group attached to the extension.